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A current-plane convention in heterodyne calibration specifies which electrical current an inferred beat amplitude denotes: the detector's generated small-signal current, the current entering a cable, or the current dissipating power in a particular load. Those amplitudes need not be equal. A power-to-current conversion is meaningful only with its circuit and reference plane.

An original photodiode model uses generated peak-current phasor $I_d$, detector shunt admittance $Y_d$ and load impedance $Z_L$. Kirchhoff's current balance givesThe factor one half assumes peak phasors; with RMS phasors it disappears. The current in the load determines delivered electric power. The generated current is recovered only after correcting the current division and any intervening network.

For a resistive Norton model with source resistance and load resistance both $R$, $Y_d=1/R$, so $I_L=I_d/2$ and $P_L=|I_d{|}^{2}R/8$. If the source shunt admittance is negligible, $I_L\approx I_d$ instead. Calling both circuits “a detector feeding a 50-ohm instrument” does not establish the same current conversion.

Capacitive current division in heterodyne receivers adds frequency dependence. With $Y_d=i\omega C_d$ and resistive load $R$,For stipulated $C_d=1$ pF and $R=50\Omega$, this magnitude is approximately 0.9540 at 1 GHz and 0.3033 at 10 GHz. The same generated amplitude then gives a smaller measured load current at the higher frequency. These are equivalent-circuit calculations, not measured specifications of a particular photodiode.

NIST's primary optical-receiver experiment distinguishes microwave power delivered to a reference load from the power indicated by a sensor. It corrects sensor calibration, interconnecting-network loss and impedance mismatch using measured scattering and reflection parameters.^[ref] That apparatus makes the reference plane an experimental calibration choice rather than a label applied after analysis.

Document the detector model, load, frequency, cable and connector corrections, and peak-versus-root mean square convention before normalizing by separate arm currents. If the arm currents describe generated DC photocurrent while the beat numerator describes uncorrected load current, the resulting overlap includes electrical current division. Heterodyne amplitude-versus-power gain conversion and impedance matching address different parts of that chain and both may be needed.

A heterodyne RF-to-DC gain transfer check tests whether the calibration converting a beat-frequency measurement into current amplitude is compatible with the calibration of the separate arm currents. The gain-matching condition for heterodyne normalization involves amplitude gains, so an RF power gain cannot be substituted without conversion.

For corrected peak beat current $B$ and arm currents $I_1,I_2$, the normalized overlap is $\gamma =B/(2\sqrt{I_1{I}_2})$. Suppose an electrical model gives sinusoidal peak current $B$ through a known resistive load $Z$. Its average beat power isSource/load division and impedance corrections must already be included when relating that load current to the detector-current convention. A factor converting power is therefore square-rooted when converting amplitude.

Let measured RF power carry multiplicative factor $g_P$, while the arm currents carry factors $g_1,g_2$. Under this model,Cancellation requires $g_P=g_1{g}_2$. With one common DC current factor $g_{\mathrm{DC}}$, the condition is $g_P=g_{\mathrm{DC}}^2$, not $g_P=g_{\mathrm{DC}}$.

An original example takes $g_{\mathrm{DC}}=1.10$. An RF power factor 1.21 produces amplitude factor 1.10 and cancels in the normalized overlap. Applying RF power factor 1.10 instead leaves multiplier $1/\sqrt{1.10}\approx 0.95346$, despite using the same numerical gain label in both paths. The example isolates heterodyne amplitude-versus-power gain conversion, rather than a change of optical overlap.

NIST's receiver experiment measured microwave power and separately corrected sensor calibration, electrical loss and impedance matching. Its two-laser optical source used polarization and spatial matching to establish calculable modulation.^[ref] Those separate operations make the RF-to-DC comparison a physical calibration problem.

Record peak versus RMS conventions, frequency response, load model and arm-current calibration. Test a known or independently characterized modulation while varying power within the linear range. Changed polarization, aperture clipping or optical alignment can alter the true overlap and imitate a gain-transfer defect, so one unexplained normalized ratio does not uniquely diagnose the electrical path.

Impedance bridging, also called voltage bridging, connects a source to a much higher load impedance so that little signal electric potential difference is lost across the source's output impedance. It is useful when the receiver should observe a signal while drawing little electric current from its source.

Represent a linear source at one frequency by open-circuit phasor $V_s$ and series electrical impedance $Z_s$. With receiving input impedance $Z_L$, the voltage divider givesThus $|Z_s/Z_L|\ll 1$ makes $H$ close to one as a complex number, controlling both amplitude and phase loading. For positive electrical resistances, the relative voltage loss is $R_s/(R_s+R_L)$. A $1\mathrm{k\Omega }$ source feeding $99\mathrm{k\Omega }$ retains $99\%$ of its open-circuit voltage; feeding $9\mathrm{k\Omega }$ retains $90\%$.

For a tolerance $\eta$ on that resistive relative loss,The often convenient ten-to-one ratio is therefore a chosen accuracy compromise, not a universal definition. A one-percent limit requires at least a ninety-nine-to-one ratio in this ideal resistive model. Voltage divider loading error gives the related loading problem when the source is itself a divider.

Impedance bridging with capacitive input explains why a large DC electrical resistance is not enough at every frequency. With source resistance $R_s$ and receiving $R_{\mathrm{in}}$ in parallel with capacitance $C_{\mathrm{in}}$,At $R_s=1\mathrm{k\Omega }$ and $R_{\mathrm{in}}=1\mathrm{M\Omega }$, the DC ratio is about $0.999$. With $C_{\mathrm{in}}=100\mathrm{pF}$, its magnitude at $1\mathrm{MHz}$ is only about $0.846$. The capacitive path loads the source despite the high DC input resistance.

Texas Instruments' three-amplifier presentation motivates high-input-impedance buffering by explicitly showing that source current and nonzero source output impedance degrade accuracy.^[ref] A buffer changes the load seen by the signal source; its own finite input and output behavior still matter.

Impedance matching has a different objective in maximum-power-transfer applications. For a resistive source, choosing equal source and load resistances gives half the open-circuit voltage, whereas bridging seeks a ratio near one. Signal fidelity, delivered power and high-frequency transmission behavior require their appropriate models.

The load impedance $Z_L$ is the terminal electrical impedance of the receiving electrical load, at a stated frequency and operating condition. With electric current defined as entering the load's positive-voltage terminal, sinusoidal steady-state phasors obeyThe complex number records both the amplitude ratio and the phase difference. It belongs to the selected terminals and model; an arbitrary nonlinear device need not have a single amplitude-independent impedance.

A resistor gives $Z_L=R$. A series resistor and capacitor giveFor $R=1\mathrm{k\Omega }$, $C=1\mu \mathrm{F}$ and $f=1\mathrm{kHz}$, this is about $(1000-159.15\mathrm{i})\Omega$, with magnitude $1012.59\Omega$ and phase $-9.04^{\circ }$. Load electric current leads its terminal electric potential difference in this example. The component laws and phasor interpretation follow the sinusoidal circuit model.^[ref]

The load changes the source's operating point. A source with open-circuit phasor $V_s$ and series impedance $Z_s$ suppliesFor RMS phasors, the complex power entering the load is $S_L=V_L{I}_L^{\ast }$. Its real part is the mean delivered power,Using peak rather than RMS amplitudes would introduce a factor $1/2$ in that power expression.

An original resistive example has $V_s=1\mathrm{V}$ RMS and $R_s=1\mathrm{k\Omega }$. A $9\mathrm{k\Omega }$ load draws $0.100\mathrm{mA}$, receives $0.900\mathrm{V}$ and dissipates $0.090\mathrm{mW}$. A $1\mathrm{k\Omega }$ load instead receives $0.500\mathrm{V}$ and dissipates $0.250\mathrm{mW}$. More delivered power need not mean a smaller voltage error.

The input impedance of an instrument is a load when the instrument is connected to a source. High-input-impedance buffering reduces this interaction in voltage measurements.^[ref] Impedance bridging prioritizes preserving the source signal; impedance matching addresses other objectives such as maximum power transfer under its assumptions.

The Johnson-Nyquist noise formula gives the open-circuit mean-square voltage in a small frequency band $d\nu$. When a source electrical resistance $R$ drives a load $R$ with impedance matching, the load sees half the source voltage, so the available noise power isThe factor of four in the open-circuit formula is therefore not an extra factor in the delivered power.

The transmitter's antenna gain $G$ makes the axial spectral energy flux at range $d$Multiplication by the receiving antenna effective area $A$ gives $d{P}_r=k_B{T}_{t}GAd\nu /(4\pi d^2)$. Define the equivalent receiving antenna temperature $T_A$ by the available matched-load noise $d{P}_r=k_B{T}_{A}d\nu$. ThenThis is the transmitter-induced equivalent noise temperature of the receiver port. It does not determine the actual physical temperature of its internal electrical resistance, or any additional receiver and background noise. The calculation assumes the usual far-field link, negligible path loss beyond geometric spreading, matched polarization of an electromagnetic wave at the two antennas and the stated effective collecting area. No additional one-half polarization factor is needed for the matched polarization of radiation from one transmitting antenna.

For electrical resistance $R_p>0$ in a parallel circuit with inductance $L>0$, add electrical admittances and invert:Conjugate impedance matching to the series RC impedance requiresIt is possible at a finite positive angular frequency exactly when $C{R}_p^2>L$, in which caseThe expression for $R$ follows by substituting this ${\omega }^2$ into the real part of $Z$. If $C{R}_p^2\le L$, no such finite matching angular frequency exists. Under literal equality, both passive inductive configurations fail by the sign of their imaginary parts.

Let the unknown series circuit contain electrical resistance $R_s>0$ and inductance $L>0$. Its electrical impedance is $Z=R_s+i\omega L$. Thus $A=R_s$, $B=L$, and conjugate impedance matching is possible withAt this angular frequency, the inductor and load capacitor have opposite electrical reactances of equal magnitude. Literal equality of their separate electrical impedances is impossible; for example, $R_s=L=C=1$ gives $1+i\omega$ versus $R-i/\omega$ for every $\omega >0$.

Use the $e^{i\omega t}$ phasor convention and interpret impedance matching as conjugate matching for the maximum power transfer theorem. The load electrical impedance of the series circuit isFor $Z=A+i\omega B$, the condition $Z_{RC}=Z^{\ast }$ equates the real parts and imaginary parts:These formulas require $B>0$, $C>0$ and a passive realization with $A\ge 0$, evaluated at the matching angular frequency; if $A,B$ depend on $\omega$, the equation for $\omega$ is implicit.

There is a convention ambiguity in the word “matched”. Literal equality $Z_{RC}=Z$, as required for equal bridge arms in part (b), instead gives $R=A$ and ${\omega }^{2}BC=-1$, requiring $B<0$. The inductive cases below use conjugate matching; literal equality is impossible for either passive inductive configuration at a finite positive angular frequency because its electrical reactance is positive while that of the series RC impedance is negative.

The characteristic impedance of a lossless transmission line is the ratio of the voltage to the current amplitude of a single forward-travelling wave. If $\mathcal{L}$ and $\mathcal{C}$ denote inductance and capacitance per unit length, the lossless telegrapher equations giveA backward wave has $I_{-}=-V_{-}/Z$ when current is positive along $+z$. Thus characteristic impedance is not a resistor's electrical resistance: it describes travelling-wave amplitudes and determines impedance matching, power transfer and reflections. A semi-infinite line, or a line terminated by a matching load, presents that impedance at its input.

At a junction with no lumped branch, voltage and electric current are continuous:Eliminating $V_t$ gives the voltage reflection coefficientFor real positive $Z_1$, incident and reflected waves occupy the same line, so their power ratio is $|r{|}^2$.

For the slightly mismatched coaxial cables, the change in characteristic impedance isConsequentlyThe requested power limit $0.01\%=10^{-4}$ implies $|\Delta b|/b<0.02\ln (b/a)$. Since $\ln (b/a)=2\pi nZ/Z_{\mathrm{vac}}=1.918$ for the stated line, the adjacent cables' outer radii must match within approximately $3.84\%$. If a tolerance is quoted separately as a symmetric manufacturing range $\pm \delta b$ and arbitrary worst-case pairs can differ by $2\delta b$, the corresponding individual tolerance is approximately $\pm 1.92\%$. The leading calculation is for small mismatches; the exact logarithmic expression gives slightly asymmetric positive and negative limits. The vacuum wave impedance used here is approximately $376.73\Omega$.^[ref]

A needle is a shunt electrical resistance $R_n$, so the load seen at its location is the continuation of the line in parallel with $R_n$: $Z_L=Z{R}_n/(Z+R_n)$. The local voltage reflection coefficient and transmission amplitude areFor periodic resistive loading of a transmission line, the reflection from the next needle travels an additional round trip $2L$ and crosses each previous needle twice. Taking the first needle as phase reference, the single-return approximation for periodic line reflections therefore sums toHere $v_p$ is the propagation phase velocity in the cable, $c/n$ for the stated nondispersive nonmagnetic filling. Squaring the amplitude gives the requested expressionFor a finite number $N$ of needles the amplitude has the additional numerator $1-[t^2{e}^{i\varphi }{]}^N$. The displayed semi-infinite result assumes sufficiently many needles, no terminal reflection, and neglects intrinsic propagation loss.

For real $r<0$, write $q=t^2$. The denominator of the power coefficient is $1+q^2-2q\cos \varphi$. It has constructive peaks at $\varphi =2\pi j$ and minima at $\varphi =(2j+1)\pi$, repeating with frequency spacing $\Delta \nu =v_p/(2L)$. Within this approximation,Weak individual shunts thus give narrow interference peaks, with phase width of order $|r|$.

The requested sum excludes paths that reflect repeatedly between different needles. It should not be read as an exact multiple-scattering result, particularly at the coherent peaks. For example the exact semi-infinite amplitude satisfiesAt $\varphi =2\pi j$ this equation reduces to $(R+1{)}^2=0$: the exact ideal infinite array has unit power reflection there, rather than the first-return approximation's value near $1/4$. This does not change the requested geometric-series derivation, but identifies its physical limitation.

An electrical transformer transfers alternating electrical power between windings through a shared changing magnetic flux. A core with large magnetic permeability confines most of the flux, so both windings link approximately the same flux $\Phi (t)$. Faraday's law gives induced electromotive forcesFor an ideal electrical transformer, negligible winding resistance and leakage flux imply the terminal-voltage magnitude ratioThe relative phase depends on the winding sense and terminal convention. At a loaded secondary, Lenz's law makes the secondary current oppose the flux change. The primary draws an additional current which approximately cancels the secondary magnetomotive force: $N_p{I}_p\simeq N_s{I}_s$ in magnitude when the magnetizing current is negligible. Consequentlyfor a resistive load, using root mean square values. More generally the ideal device conserves instantaneous power, with the appropriate current signs. A load impedance $Z_L$ appears at the primary as $Z_{\mathrm{in}}=(N_p/N_s{)}^2{Z}_L$, enabling impedance matching and electrical isolation.

For sinusoidal flux, $\Phi ={\Phi }_{\max }\sin \omega t$ gives $V_{\mathrm{rms}}=\omega N{\Phi }_{\max }/\sqrt{2}$. At a given supply voltage, reducing frequency increases the required peak flux and can drive the core into magnetic saturation. Real losses include winding Joule heating, magnetic hysteresis, eddy currents and leakage flux; a laminated core suppresses eddy-current loops. A steady direct current produces no continuing secondary electromotive force, while its small winding resistance can allow a damaging primary current after switching transients.